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# Calculating flow through 2 orifices5

## Calculating flow through 2 orifices

(OP)
Hi everyone,

Lets say you have a small hydraulic motor spinning at 1500rpm at 8l/min with no restriction. Then I introduce a 1.3mm restriction on one side and it makes no difference, but if I add another on the other side it slows to 1200rpm. Can someone explain how another restrictor of the same size can have such a dramatic effect on the speed?

If posting any formulas, metric would be idea.

Thanks

### RE: Calculating flow through 2 orifices

What is the power source? What are the controlling devices? Have you measured pressures of supply and in and out? What other control devices, if any, are in the circuit? sequence, flushing, counterbalance, etc.

Assuming the motor leakage is not an issue, I would guess you have a constant volume (fixed gear or similar) pump, and the circuit was constant flow originally. Pressure increased with first orifice, still holding constant flow. With addition of the second orifice, the circuit changed to constant pressure because the increased load put the required prssure above the relief valve setting, and the excess flow is going across relief.

If you have a constant pressure variable pump circuit, then there may be leakages. Or, the pump may have been at max flow, acting as a contant flow device until the increased load put it on the compensator.

Excluding leakages and other quirks unknown.

kcj

### RE: Calculating flow through 2 orifices

As kcj said, the second orifice raised the system pressure up to or above some controlling component pressure setting and flow decreased. If the motor does not have a case drain for the shaft seal, you run the risk of overpressuring the shaft seal by putting an orifice downstream of the motor.

More circuit and component information would be helpful.

Ted

### RE: Calculating flow through 2 orifices

(OP)
Fixed gear pump with a fixed speed motor. The function is operated by a standard spool. I don't have the info on the exact type of spool used. I understand the flow must be going somewhere else, that isn't the issue.

The main thing I am confused by is how 2 orifices have such a dramatic effect. Each restrictor is a male to male fitting with a little brass insert with a drilled hole about 3mm in length. If this was 6mm in length, would it have the same effect? If not, why does one either side do this?

Thanks

### RE: Calculating flow through 2 orifices

C'mon Hydroman247, do you seriously believe that the insertion of a single 1.3 mm orifice in your 8 L/min flow is having absolutely no effect on anything in your system? And that there has to be two such orifices inserted before there is any effect at all?

Of course the single orifice is having an effect; you're just not looking in the right place. Then, when you put two such orifices in the circuit, that original effect should be "doubled", except that the combined effect of the two orifices seems to have activated something else in your circuit (relief valve opens, reducing valve closes, pump displacement starts to reduce etc.) The reduction in motor speed is the only thing you see because it's possibly the only thing you're looking at.

If we treat the restriction as a sharp edged orifice with a diameter of 1.3 mm and a discharge coefficient of 0.8, then a flow of 8 L/min with a fluid density of 880 kg/m³ gives a calculated pressure drop of 70 bar.

If we treat the restriction as a 3 mm length of 1.3 mm bore tube with a fluid viscosity of 32 cSt then we would calculate a pressure drop of 40 bar.

Neither of these calculations is perfect because your restriction is too long to be treated as "sharp edged" and isn't long enough to be treated as a "tube". If you double the length of the 1.3 mm drilling to 6 mm then it becomes more tube like and less orifice like - but the simplistic calculations will still be inaccurate and you probably won't see a doubling of the original pressure drop.

So what are we left with? Your first orifice will create a particular pressure drop when there is a flow through it, say Y bar. Another identical orifice will create an identical pressure drop, another Y bar. Put two such orifices in your circuit and the pressure needed to drive the original flow through your circuit will increase by the sum of the two pressure drops, i.e., 2Y bar.

However, something non-linear has happened in your circuit - it cannot tolerate the additional restriction without the flow reducing. That's why the motor speed has dropped.

But here's another thought: some hydraulic motors react badly to pressure on their outlet port; the mechanical efficiency drops hugely. It could be that the downstream orifice is upsetting the motor in some way. Have you tried the circuit with just the downstream orifice in place?

I have to agree with Ted and kcj; you need to send more details of measured pressures, component types and sizes and the various settings of the adjustable valves etc. But I'm sort of imagining that, when you gather together this pertinent data, the answer to your question will become kind of obvious.

DOL

### RE: Calculating flow through 2 orifices

(OP)
Thanks for that. The problem is I don't have all the specs of this system here. I was just making an example. I haven't got the actual flow rates or motor rpm. I am just interested in why the effect is doubled with 2 restrictors of the same size.

If I have a tube with some sort of restriction at one end and try to blow through it, I can feel how hard it is to blow through, but if I put another restriction at the other end, it wouldn't be twice as hard to blow through. Is this purely to do with the viscosity of the fluid?

Thanks

### RE: Calculating flow through 2 orifices

(OP)
I have just found this formula for putting restrictors in series.

SQRT( (diameterA^2 * diameterB^2) / (diameterA^2+diameterB^2))

Which gives me answers like this.

0.5mm + 0.5mm = 0.35mm equivalent.
1.3mm + 1.33 - 0.92mm equivalent.

Who is correct here?

### RE: Calculating flow through 2 orifices

Your formula is not quite correct, it should be:

[(diameterA^4 * diameterB^4)/(diameterA^4 + diameterB^4)]^0.25

This means that two 1.3 mm diameter restrictors in series create the same pressure drop as a single restrictor of 1.093 mm diameter.

DOL

### RE: Calculating flow through 2 orifices

Thanks for the equation. My gut was a much lower effectice orifice area. (a hydrualic circuit I worked with was modified to replace a small orifice with two of a larger diameter with the reasoning to avoid possible trash plugging)

### RE: Calculating flow through 2 orifices

hydroman247, your sample problem does not have two orifices in series. One is before the motor and one is after the motor. Is series orifice relevant?

Ted

### RE: Calculating flow through 2 orifices

(OP)
@oldhydroman Thanks for that, updated my spreadsheet for future use.

@byrdj, when looking for formulas, this was the reason most people wanted to fit 2 restrictors in series.

@hydtools, No, but can someone explain how I would work out the required orifices and whether one would be an option. It would alternate being the input and the output.

The motor in question is bi-directional so putting a restrictor just on the exhaust port might not be an option. I understand the basic principle here, I think, it just doesn't make sense to me when I think about it.

### RE: Calculating flow through 2 orifices

Are you asking how to add flow controls to a bi-directional hydraulic motor?

If so then throttle-check valves would be the usual choice. If the load, temperature (viscosity) and the system pressure remain constant then an ordinary throttle-check valve will do. This is a common body (or cartridge) incorporating a needle valve and a check valve. Flow in one direction has to pass over the needle valve, flow in the reverse direction goes over the check valve and is not restricted.

The basic, line mounted throttle-check valve will have a block body with an adjustment knob sticking out of the top. You have to install this block on swivel fittings to get the knob orientation correct.

A simpler version has a barrel type construction, no swivel fittings are required because you turn the outer sleeve to make the flow adjustement so the valve has no particular installation orientation requirements. You can lock the setting quite effectively with the sleeve locking ring and two pipe wrenches. Once set these devices attract very little unwarranted 'attention'. Barrel type valve cannot be used as isolators and frequently cannot be adjusted while pressurized.

If you don't want the speed to be adjustable then you can fit a check valve with a built in orifice - See Sun hydraulics type CNBC.

If you want the speed to remain constant with widely varying fluid viscosity then you need a temperature compensated throttle-check valve: basically the flow control function is formed from two "sharp" edges arranged to move together so that the adjustable restriction is effectively "zero length".

If you want the speed to remain constant with varying load and/or varying system pressure then you will need a "pressure compensated" flow control valve (with reverse free-flow check). This type of valve incorporates a hydrostat to keep a constant pressure drop across the adjustable flow control part. Don't pick this valve if you really don't have to because you will be wasting your money, there's more to go wrong, there is a minimum pressure drop needed to get it to function and it usually starts with a jump (there are ways of overcoming the jump but your system might not need that degree of complication).

If you want this pressure compensation feature with widely varying viscosities as well then you will need a "pressure compensated and temperature compensated flow control valve with reverse free-flow check". [PLEASE - someone invent a shorter name!]

You also need to decide if the flow controls will be installed in a meter-in or a meter-out configuration. It is possible to have meter-in in one direction and meter-out on the other direction.

DOL

### RE: Calculating flow through 2 orifices

(OP)
Thanks for all that info. I am currently working on another solution that involves changing the speed of the motor for other reasons and may be able to tune the speed that way.

This machine will be operating in various climates and different loads but so long as the speed of the function this motor operates falls within certain parameters then there is no issue. We had issues with other functions so sped the motor up and it upset the speed of this one function but I may be able to find a mid point which can satisfy both needs.

The other problem with mobile is that adding multiple check valves and other valves is more of an issue than it would be in a static system. Simple restrictors are small and cheap.

My question is why is a restrictor either side of a motor, any different to 2 restrictors in series?
I understand the motor creates a pressure drop as well so effectively 3 restrictors but the end result will be the same wouldn't it?

### RE: Calculating flow through 2 orifices

Based on your results with produces, I will assume you have a constant flow delivery. Use a bypass, three port, pressure and temperture compensated flow control with built-in reverse flow check valve. One on each side of the motor. Controlled flow arrow pointing to the motor.

Ted

### RE: Calculating flow through 2 orifices

Plumb the bypass flow directly back to tank. Put the flow control before the directional valve and you need only one flow control.

Ted

### RE: Calculating flow through 2 orifices

(OP)
Thanks hydtools but that problem is semi solved. Will still give you a star. :D

My question is why is a restrictor either side of a motor, any different to 2 restrictors in series?
I understand the motor creates a pressure drop as well so effectively 3 restrictors but the end result will be the same wouldn't it?

### RE: Calculating flow through 2 orifices

Yes. Same flow through all three devices, same pressure drop across each no matter the order in series.

Ted

### RE: Calculating flow through 2 orifices

(OP)
Ok, thanks. Will be back in a few weeks with another question no doubt.

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